Physics MCQs for NEET — Practice Questions with Answers

For vertically upward motion, $h_1=v_{0}t-\frac{1}{2}g{t}^2$ and for vertically down ward motion, $h_2=v_{0}t+\frac{1}{2}g{t}^2$

∴ Total distance covered in t sec $h=h_1+h_2=2v_{o}t$. 

Let height of tower is h and body takes t time to reach to ground when it fall freely.

∴ $h=\frac{1}{2}g{t}^2$ …(i)

In last second i.e. t^th sec body travels = 0.36 h

It means in rest of the time i.e. in (t – 1) sec it travels

$=h-0.36h=0.64h$

Now applying equation of motion for (t – 1) sec

$0.64h=\frac{1}{2}g{(t-1)}^2$ …(ii)

From (i) and (ii) we get, t = 5 sec and h = 125 m

$$\begin{gathered} After time t_1: \\ Time to come back to height h after time t_1=t_2-t_1 \\ Time to reach highest point from height h=time to come back to height h from highest point=\frac{t_2-t_1}{2} \\ Time to reach highest point from ground=\frac{t_2-t_1}{2}+t_1=\frac{t_1+t_2}{2} \\ \Rightarrow \frac{t_1+t_2}{2}=\frac{u}{g} \\ \Rightarrow u=\frac{g{\left(t_1+t\right)}_2}{2} \end{gathered}$$

For first projectile, $h_1=ut-\frac{1}{2}g{t}^2$

For second projectile, $h_2=u(t-T)-\frac{1}{2}g{(t-T)}^2$

When both meet i.e. $h_1=h_2$

$ut-\frac{1}{2}g{t}^2=u(t-T)-\frac{1}{2}g{(t-T)}^2$

⇒ $uT+\frac{1}{2}g{T}^2=gtT$

⇒ $t=\frac{u}{g}+\frac{T}{2}$

and $h_1=u\left(\frac{{\displaystyle u}}{{\displaystyle g}}+\frac{{\displaystyle T}}{{\displaystyle 2}}\right)-\frac{1}{2}g{\left(\frac{{\displaystyle u}}{{\displaystyle 2}}+\frac{{\displaystyle T}}{{\displaystyle 2}}\right)}^2$

$=\frac{u^2}{2g}-\frac{g{T}^2}{8}$

(d) Velocity of each car is given by -

$v_p=\frac{dx\left(t\right)}{dt}=a+2bt$

$\mathrm{And}, {\mathrm{v}}_{\mathrm{Q}}=\frac{{\mathrm{dx}}_{\mathrm{Q}}\left(\mathrm{t}\right)}{\mathrm{dt}}=\mathrm{f}-2\mathrm{t}$

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that}- \\ {\mathrm{v}}_{\mathrm{P}}={\mathrm{v}}_{\mathrm{Q}} \\ \Rightarrow \mathrm{a}+2\mathrm{bt}=\mathrm{f}-2\mathrm{t} \\ \Rightarrow \mathrm{t}=\frac{\mathrm{f}-\mathrm{a}}{2\left(\mathrm{b}+1\right)} \end{gathered}$$ 

Given, $x=8+12t-t^3$

We know $v=\frac{dx}{dt}$

So,     $v=12-3t^2$

When v=0, t=2 sec

and     $a=\frac{dv}{dt}$

and      a=-6t

At         t=2s,

                  $a=-12m/s^2$

So, retardation of the particle = 12m/$s^2.$

Velocity $\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}}\Rightarrow \mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}$

 The average speed of the particle :

        $$\begin{gathered} {\mathrm{v}}_{\mathrm{av}}=\frac{\mathrm{s}+\mathrm{s}}{\frac{\mathrm{s}}{{\mathrm{v}}_1}+\frac{\mathrm{s}}{{\mathrm{v}}_2}} \\ {\mathrm{v}}_{\mathrm{av}}=\frac{2{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{v}}_1+{\mathrm{v}}_2} \end{gathered}$$

(c)

Key Idea    If the particle is moving in a straight line under the action of a constant force then

   distance covered $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2.$

   Since the body starts from rest u=0 

$$\begin{gathered} \therefore \mathrm{s}=\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{Now}, {\mathrm{s}}_1=\frac{1}{2}\mathrm{a}{\left(10\right)}^2 ....\left(\mathrm{i}\right) \\ \mathrm{and} {\mathrm{s}}_2=\frac{1}{2}\mathrm{a}(20{)}^2 ....(\mathrm{ii}) \\ \mathrm{Dividing} \mathrm{Eq}. (\mathrm{i}) \mathrm{and} \mathrm{Eq}. (\mathrm{ii}), \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}=\frac{(10{)}^2}{(20{)}^2} \\ \Rightarrow {\mathrm{s}}_2=4{\mathrm{s}}_1 \end{gathered}$$

(c)

Distance travelled by the particle in nth second is 

    $S_{nth}=u+\frac{1}{2}a\left(2n-1\right)$

where u is initial speed and a is acceleration of the particle.

Here, n=3, u=0, a=$\frac{4}{3}m/s^2$

$\therefore S_{3rd}=0+\frac{1}{2}\times \frac{4}{3}\times \left(2\times 3-1\right)$

                 = $\frac{4}{6}\times 5$

                =$\frac{10}{3}m$

Alternatively : Distance travelled in the 3rd second = distance travelled in 3s - distance travelled in 2s

As, u=0,

 $S_{\left(3rd s\right)}=\frac{1}{2}a.3^2-\frac{1}{2}a.2^2=\frac{1}{2}.a.5$

$Given a=\frac{4}{3}m{s}^{-2}$

  $\therefore S_{\left(3rd s\right)}=\frac{1}{2}\times \frac{4}{3}\times 5=\frac{10}{ 3}m$

The problem requires kinematics equations of motion. 

Let u and v  the first and final velocities of particle and a and s be the constant acceleration and distance covered by it. 

from third equation of motion 

        $$\begin{gathered} v^2=u^2+2as \\ \Rightarrow {\left(20\right)}^2={\left(10\right)}^2+ 2a\times 135 \\ or a=\frac{300}{2\times 135}=\frac{10}{9}m{s}^{-2} \end{gathered}$$

Now using first equation of motion,

                 v=u+ at

or     t=$\frac{v-u}{a}=\frac{20-10}{\left(10/9\right)}=\frac{10\times 9}{10}=9s$